Question

Let (m,n) be integer solutions to: 7m+12n = 22

What is the greatest negative resulting from: m+n?

Answer 1

im assuming a greatest negative to be the closest to zero

Answer 2

Solve m in terms of n and n in terms of m, I think it will help you.

Answer 3

Then find what integer values for one will make the other an integer.

Answer 4

these are the norms!

Answer 5

hmm, i did a few euclidian runs to find a particular solution; then used that to determine the general solutions for m and n.

Answer 6

man its difficult.. sorry idk how to solve this coplex prob.

Answer 7

but I'll try once

Answer 8

your suggestion would start out:\[m=\frac{22-12n}{7}\]\[n=\frac{22-7m}{12}\]
\[m+n=...\]
doesnt really get us an equation in 1 variable

Answer 9

my steps:

find integer solutions to: 7m+12n = gcd(12,7)

7m+12n = 1

the euclidean algorithm helps us out

12 = 7(1)+5
7 = 5(1)+2
5 = 2(2)+1

the reverse euclidean algorithm lets us define specific m or n

0 -1( 1) = -1
1 -1(-1) = 2
-1 -2( 2) = -5 , let m=-5, n=3

7(-5) + 12(3) = 1 ; multiply thru by 22

7(-110) + 12(66) = 22 ; (-110,66) are particular solutions to the problem

Answer 10

let the particular solution be (m',n')

we can determine equations for m and n as follows:

m = m' + 12k

n = n' - 7k

Answer 11

I think tazz correct, but wht does the answer book say,

Answer 12

i havent chked the answer key yet

Answer 13

chk that I am sure about it that u r correct

Answer 14

m+n = m'+n' + 5k

-110 + 66 + 5k <= 0

5k <= 44

k <= 8.xxxx , so try k=8

-44 + 5(8) = -4

Answer 15

im hoping some alternative methods surface :) i like seeing different ways to approach a problem

Answer 16

u can find general solution for m and n in terms of one variable\[m=\frac{22-12n}{7}=3-n+\frac{1-5n}{7}\]so \(5n-1=7k\) in a similar process\[n=\frac{7k+1}{5}=k+\frac{2k+1}{5}\]and again \(2k+1=5t\) so\[k=\frac{5t-1}{2}=2t+\frac{t-1}{2}\]and finally \(t=2s+1\) so we can write m and n in terms of s, i hope its right :)

Answer 17

sub back one by one to get\[k=5s+2\]\[n=5s+2+2s+1=7s+3\]\[m=3-7s-3+\frac{1-45s-15}{7}=-7s-5s-2=-12s-2\]finally \((m,n)=(-12s-2,7s+3)\)

Answer 18

i like the division processes, which looks to relate to the 7m+12n = 1

Answer 19

-12s-2
7s+3
--------
-5s + 1 <= 0

s>=1/5; take s=1

we still get to -4 :)

Answer 20

:)